Let $n$ be a natural number. The HCF of $n, n + 10$ is 10. If the LCM is $x$ (a 2-digit number), then how many values of $x$ are possible?

Question

Prepp Team · Accepted Answer

Analyzing HCF and LCM for n, n+10

We are given a problem involving the Highest Common Factor (HCF) and Least Common Multiple (LCM) of two numbers, $n$ and $n + 10$. We know that $n$ is a natural number. The HCF of $n$ and $n + 10$ is given as 10. Their LCM is represented by $x$, and we are told $x$ is a 2-digit number. Our goal is to determine how many different values $x$ can possibly take.

Essential HCF and LCM Properties

To solve this, we'll use some basic but important properties of HCF and LCM:

HCF Definition: The HCF of two or more numbers is the largest positive integer that divides each of the numbers without leaving a remainder.
LCM Definition: The LCM of two or more numbers is the smallest positive integer that is a multiple of each of the numbers.
HCF-LCM Product Rule: For any two positive integers $a$ and $b$, the product of the numbers is equal to the product of their HCF and LCM. Mathematically, this is written as:
HCF($a, b$) $\times$ LCM($a, b$) = $a \times b$
HCF Representation: If HCF($a, b$) = $d$, then we can express $a$ and $b$ as multiples of $d$. Specifically, $a = d \cdot p$ and $b = d \cdot q$, where $p$ and $q$ are integers that are co-prime (meaning HCF($p, q$) = 1).

Solving the HCF LCM Problem

Let's break down the problem step-by-step using these properties.

Step 1: Using HCF to Define n and n+10

We are given HCF($n, n + 10$) = 10. Applying the HCF representation property:

$n = 10p$

$n + 10 = 10q$

Here, $p$ and $q$ must be co-prime integers (HCF($p, q$) = 1). Since $n$ is a natural number ($n \ge 1$), $p$ must also be a positive integer ($p \ge 1$).

Step 2: Deriving Relationship Between p and q

Let's find the relationship between $p$ and $q$. We can substitute the expression for $n$ from the first equation into the second equation:

$ (10p) + 10 = 10q $

Now, divide the entire equation by 10:

$ p + 1 = q $

This result shows that $q$ is always exactly one greater than $p$. Consecutive integers ($p$ and $p+1$) are always co-prime. Thus, HCF($p, p+1$) = 1 is automatically satisfied for all positive integers $p$.

Step 3: Formula for LCM (x)

Now, let's use the HCF-LCM Product Rule: HCF($n, n + 10$) $\times$ LCM($n, n + 10$) = $n \times (n + 10)$.

Substituting the known values and expressions:

$ 10 \times x = (10p) \times (10q) $

$ 10x = 100pq $

Substitute $q = p+1$ into this equation:

$ 10x = 100p(p+1) $

To find $x$, divide both sides by 10:

$ x = 10p(p+1) $

Step 4: Finding Possible Values of x

The problem states that $x$ must be a 2-digit number. This means $x$ must be between 10 and 99, inclusive.

$ 10 \le x \le 99 $

Substitute the formula we found for $x$:

$ 10 \le 10p(p+1) \le 99 $

Divide all parts of the inequality by 10:

$ 1 \le p(p+1) \le 9.9 $

We need to find the positive integer values of $p$ (starting from $p=1$) that satisfy this condition.

Testing p = 1: $p(p+1) = 1(1+1) = 1(2) = 2$. Is $1 \le 2 \le 9.9$? Yes. The LCM value is $x = 10 \times 2 = 20$. This is a 2-digit number.
Testing p = 2: $p(p+1) = 2(2+1) = 2(3) = 6$. Is $1 \le 6 \le 9.9$? Yes. The LCM value is $x = 10 \times 6 = 60$. This is also a 2-digit number.
Testing p = 3: $p(p+1) = 3(3+1) = 3(4) = 12$. Is $1 \le 12 \le 9.9$? No, because $12 > 9.9$. The LCM value would be $x = 10 \times 12 = 120$, which is a 3-digit number and thus not possible.

For any integer $p > 2$, the value of $p(p+1)$ will be greater than 9.9, so no further values of $p$ will yield a 2-digit LCM.

Step 5: Counting the Possible x Values

We found that only $p=1$ and $p=2$ result in a 2-digit LCM ($x$).

The possible values for $x$ are 20 and 60.

Therefore, there are exactly two possible values for $x$.

Conclusion on Possible x Values

The analysis shows that there are two possible values for $x$, the 2-digit LCM, under the given conditions.

Prepp Team · Answer

Only one

Prepp Team · Answer

Only three

Prepp Team · Answer

More than three

Analyzing HCF and LCM for n, n+10

Essential HCF and LCM Properties

Solving the HCF LCM Problem

Step 1: Using HCF to Define n and n+10

Step 2: Deriving Relationship Between p and q

Step 3: Formula for LCM (x)

Step 4: Finding Possible Values of x

Step 5: Counting the Possible x Values

Conclusion on Possible x Values

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